Question

A can hit a target 4 times in 5 shots, B can hit 3 times in 4 shots and C twice in 3 shots. They fire a volley. What is the probability that two shots hit the target?
A. $\dfrac{13}{30}$
B. $\dfrac{17}{30}$
C. $\dfrac{11}{30}$
D. None of these.

Answer 1

Hint: To solve this question, we need to know the concept of probability. We know that probability is the ratio of the number of favourable outcomes to the total number of outcomes. Mathematically, we can write it as, $\text{Probability =}\dfrac{\text{Favourable outcomes}}{\text{Total number of outcomes}}$.

Complete step-by-step answer:

In this question, we are asked to find the probability of two shots hitting the target. To find this, we should know the concept of probability. Probability is nothing but the ratio of number of favourable outcomes to the total number of outcomes. Mathematically, we can write it as, $\text{Probability =}\dfrac{\text{Favourable outcomes}}{\text{Total number of outcomes}}$.
In this question, we are taking the probability of hitting the target. So, the ratio of the number of times that the target is being hit to the total number of attempts made to hit the target will give us the probability of hitting the shot. So, we can say that,
Probability of A hitting the target = P (A) = $\dfrac{4}{5}$.
Probability of B hitting the target = P (B) = $\dfrac{3}{4}$.
Probability of C hitting the target = P (C) = $\dfrac{2}{3}$.
Now, in this question, we have to find the probability of two shots hitting the target, that means, any one of A, B or C misses the shots.
Now let us consider that A misses the shot, then the probability of B and C hitting the target is, $P\left( \bar{A}BC \right)=\dfrac{1}{5}\times \dfrac{3}{4}\times \dfrac{2}{3}\Rightarrow \dfrac{1}{10}$.
Now let us consider that B misses the shot, then the probability of A and C hitting the target is, $P\left( A\bar{B}C \right)=\dfrac{4}{5}\times \dfrac{1}{4}\times \dfrac{2}{3}\Rightarrow \dfrac{2}{15}$.
Now let us consider that C misses the shot, then the probability of A and B hitting the target is, $P\left( AB\bar{C} \right)=\dfrac{4}{5}\times \dfrac{3}{4}\times \dfrac{1}{3}\Rightarrow \dfrac{1}{5}$.
Now, the total probability of 2 shots hitting the target can be found out by adding all the previous probabilities. So, it can be written as:
Probability of 2 shots hitting the target = $P\left( \bar{A}BC \right)+P\left( A\bar{B}C \right)+P\left( AB\bar{C} \right)$.
$\begin{align}
  & \Rightarrow \dfrac{1}{10}+\dfrac{2}{15}+\dfrac{1}{5} \\
 & \Rightarrow \dfrac{3+4+6}{30} \\
 & \Rightarrow \dfrac{13}{30} \\
\end{align}$
Therefore, the probability of 2 shots hitting the target is $\dfrac{13}{30}$. Hence, option (A) is the correct answer.

Note: The possible mistake is that students may only find any one of $P\left( \bar{A}BC \right),P\left( A\bar{B}C \right)$ and $P\left( AB\bar{C} \right)$ and then multiply it by 3, which is wrong, as the probability of A hitting the target, the probability of B hitting the target and the probability of C hitting the target are all different.